91影视

So, let's take the above function and plug it back into the Schrodinger equation to see if we can come up with a condition on the value of $\mathbf{\text{蝇}}$or $\mathit{k}$ that will allow us to get this result.

$鈭�\frac{{\mathrm{鈩�}}^2}{2m}\frac{{鈭�}^2唯}{鈭�x^2}=i\mathrm{鈩�}\frac{鈭�唯}{鈭�t}$鈥︹赌︹赌︹赌︹赌︹赌�(1)

The wavefunction given is

$唯=\text{Acos(kx-蝇t)}$鈥︹赌︹赌︹赌︹赌︹赌�.(2)

Substitute the value of wavefunction from equation (2) into equation (1), and we get,

$\begin{array}{l}\frac{{鈭�}^2唯}{鈭�x^2}=鈭�{\text{Ak}}^{\text{2}}\text{cos(kx-蝇t)}\\ \frac{鈭�唯}{鈭�t}=\text{A蝇sin(kx-蝇t)}\end{array}$

$\begin{array}{l}鈭�\frac{{\mathrm{鈩�}}^2}{\text{2m}}脳\left({\text{-Ak}}^{\text{2}}\text{cos(kx-蝇t)}\right)=i\mathrm{鈩�}脳\text{A蝇sin(kx-蝇t)}\\ \frac{\mathrm{鈩�}{\text{k}}^{\text{2}}}{\text{2m}}=\text{i蝇tan(kx-蝇t)}\end{array}$

Unlike the exponential case, this final expression has a function that depends on both position and time, but the left-hand side is constant. As a result, we can't discover a condition on $k$or $\text{蝇}$ that will make the two sides equal, and the sine function as a solution is no different.